Computing winding number Need some help with computing index number (winding number).
I need to compute the index number: $n(\gamma,0)$ where $\gamma(t)=\cos t + 3 i \sin t$ for $0 \leq t \leq 4\pi$
I tried to use parametrization, but the integral seem quite ugly. By the use of computer the answer is 0, but it can't be because the index counts the number of circuit on this ellipse, it should be 2 as i see it.
Maybe you have a way to compute it?
 A: Clearly the winding number is $2$, because $2\pi$ worth of $t$ is one circuit around the origin, and there are two of them. This argument ought to be sufficient everywhere except in specialized classroom situations where you're required to use some particular method to determine the winding number. Since you're not specifying a method I assume this is not the case for you.
If you do want some kind of "more rigorous" argument, I recommend something like:

Let's construct a continuous argument function $f(t)$ for $\gamma$. Since $\gamma(0)=1$, let's set $f(0)=0$.
For $0\le t\le \pi$ the imaginary part of $\gamma(t)$ is nonnegative, so let's make $f(t)\in[0,\pi]$ in this interval. This leads, in some way we don't need to care about, to $f(\pi)=\pi$.
For $\pi\le t\le 2\pi$ the imaginary part of $\gamma(t)$ is nonpositive, so let's make $f(t)\in[\pi,2\pi]$ in this interval. This leads, in some way we don't need to care about, to $f(2\pi)=2\pi$.
For $2\pi\le t\le 3\pi$ the imaginary part of $\gamma(t)$ is nonnegative, so let's make $f(t)\in[2\pi,3\pi]$ in this interval. This leads, in some way we don't need to care about, to $f(3\pi)=3\pi$.
... and so forth ...

A: The winding number $n(\gamma, a)$ is defined by
$$n(\gamma, a) = {1 \over 2\pi i} \int_{\gamma} {1 \over z - a}\,dz$$
So you are looking for 
$${1 \over 2\pi i} \int_{\gamma} {1 \over z}\,dz$$
The Cauchy integral formula will give that the integral over each circuit of this is $1$, so the overall answer is $2$.  
